Question 74256This question is from textbook Algebra 2
: State the number of positive real zeros, negative real zeros, and imaginary zeros for g(x)=9x^3-7x^2+10x-4.
I am getting very confused by finding the zeros of an equation. My book does not explain things clearly and the one example they give doesn't apply to the problems on the practice worksheet. Can you please help me with this and explain in detail? I really need something to work from. Thank you in advance!
This question is from textbook Algebra 2
Answer by rmromero(383)   (Show Source):
You can put this solution on YOUR website!
State the number of positive real zeros, negative real zeros, and imaginary zeros for g(x)=9x^3-7x^2+10x-4.
Descartes Rule can help you.
g(x) = 9x^3 - 7x2 + 10x - 4
+ - + - Notice the signs
the signs varies 3 time + -, - +, + - therefore there are 3 variations
That means that there are 3 possible positive real zeros or 1 positive
real zeros
Now
g(-x) = 9(-x)^3 - 7(-x)^2 + 10(-x) - 4
= -9x^3 + 7x^2 - 10x - 4
= - + - - Notice the signs
The signs varies 2 times - +, + - therefore there are 2 variations
that means that there are 2 possible negative zeros
With the given variations the possible zeros are:
3 positive real zeros, 0 negative real zeros, 0 imaginary
or
1 positive real zeros, 2 negative real zeros, 0 imaginary
Remember that the highest degree x^3 determines the number of zeros you have
so each total possible zeros are 3
I hope this help
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